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I guess this question is quite easy but I cannot prove this. Please help me.enter image description here

But I cannot understand the last step.

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    Hint: $(a-b)(a+b)=a^2-b^2\,$. – 2017-01-30
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    Are you familiar with $A^2 - B^2 = (A-B)(A+B)$? – 2017-01-30

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Well, \begin{split} |\sqrt{x} - \sqrt{x_{0}}| &= |(\sqrt{x} - \sqrt{x_{0}}) \cdot 1|\\ &= \left |(\sqrt{x} - \sqrt{x_{0}}) \cdot \dfrac{(\sqrt{x} + \sqrt{x_{0}})}{(\sqrt{x} + \sqrt{x_{0}})} \right | \\ &= \left |\dfrac{(\sqrt{x} - \sqrt{x_{0}})(\sqrt{x} + \sqrt{x_{0}})}{(\sqrt{x} + \sqrt{x_{0}})} \right | \\ &= \left |\dfrac{x - x_{0}}{\sqrt{x} + \sqrt{x_{0}}} \right | \end{split} where you can either multiply $(\sqrt{x} - \sqrt{x_{0}})(\sqrt{x} + \sqrt{x_{0}})$ out in the second to last step, or just recognize it as a product of the form $(a - b)(a+b)$, which multiplies out to $a^{2} - b^{2}$, the difference of squares.

EDIT Note that in the body of your question, your absolute value symbol disappears from the denominator. This is because $\sqrt{x}$ and $\sqrt{x_{0}}$ are both positive, and so their sum is positive. Clearly, if $a \geq 0$, then $|a| = a$.

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    Maybe you add why the modulus "magically disappears" in the denominator. – 2017-01-30
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    @TheGeekGreek Good idea, geeky. :) – 2017-01-30
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    Thanks! It was really trivial but sometimes things can look nebulous. Thanks! – 2017-01-30
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    @PedroGomes I totally understand what you mean. You're welcome. – 2017-01-30